S3 Group

The S3 group is the group of permutation on 3 elements. It has 6 elements. It is the group of all the automorphism on the set {1, 2, 3}.

Here is the full list of the elements of S3:

## e: ()
## a: (1,2)
## b: (1,3)
## c: (2,3)
## p: (1,2,3)
## p^2: (1,3,2)

Table of S3:

S3	e	p	p2	a	b	c
e	e	p	p2	a	b	c
p	p	p2	e	c	a	b
p2	p2	e	p	b	c	a
a	a	b	c	e	p	p2
b	b	c	a	p2	e	p
c	c	a	b	p	p2	e

Order of the elements:

S3	Order	Sign	Inversions
e	1	+	0
p	3	+	0
p2	3	+	0
a	2	-	1
b	2	-	1
c	2	-	1

Unique normal chain decomposition of S3:

\[ C(2) \lhd C(3) \lhd S_3\]

And this decomposition is unique, the factors cannot be swapped.

Cycle graph

grViz('
graph cycle {
 graph[layout = neato]
 node [shape = circle, style = filled, color = grey, label=""]

 e [color = red]; a; b; c; p; p2

  e -- a
  e -- b
  e -- c
  e -- p -- p2 -- e
}
')

Subgroups of S3

###

is a normal subgroup

The group <p> = {e, p, p2} is normal in S3:

all(p^a == p2, p^b == p2, p^c == p2)

## [1] TRUE

The classes of <p> in S3 are:

{e, p, p2}, {a, b, c}

The action of a on <p>: (p p2)

More detailed action of the group on <p>:

S3	{e, p, p2}	{a, b, c}
e	{e, p, p2}	{a, b, c}
a	{a, b, c}	{e, p, p2}
b	{b, c, a}	{p2, e, p}
c	{c, a, b}	{p, p2, e}
p	{p, p2, e}	{c, a, b}
p2	{p2, e, p}	{b, c, a}

<p> is isomorphic to Z/3Z and it has no non-trivial subgroups.

If G is a subgroup of S3 containing <p> then G*<p> is a subgroup of S3/<p>. S3/<p> is a group of order 2 so it is isomorphic to Z/2Z and it has no non trivial subgroup. The only subgroups containing p are {e} and S3.

If G is a sub-group not including <p>. Then inter(G, <p>) is {e}. because inter(G, <p>) is a subgroup of <p> and it cannot be <p> because G does no include <p>.

So G is included in {e, a, b, c}. G cannot contain more than 2 elements because any of products ab, bc, … will yield an element of <p>. So G is either {e, a}, {e, b}, {e, c}.

Conclusion: Here is the full list of the subgroups of S3: {e}, {e, a}, {e, b}, {e, c}, {e, p, p2}, {e, a, b, c, p, p2}

Lattice of all the subgroups

## Warning in .Call("R_igraph_layout_fruchterman_reingold", graph, coords, :
## '.Random.seed' is not an integer vector but of type 'NULL', so ignored

Normal subgroups

Actions of the group

The natural action of S3 on {1, 2, 3}

This is a faithful action.

We can compute the orbits:

S3	Orbits
e	{1} {2} {3}
a	{1, 2} {3}
b	{1, 3} {2}
c	{1} {2, 3}
p	{1, 2, 3}
p2	{1, 2, 3}

We can also compute the stabilizers:

{1, 2, 3}	Stabilizer
1	<c>
2	<b>
3	<a>

Action on the ‘edges’

We consider the action of S3 on this set: {{1, 2}, {1, 3}, {2, 3}}

• Action of <p>

## [1] TRUE

• Action of a:

all(
  a_f(c(1, 2)) == c(2, 1),
  a_f(c(1,3)) == c(2, 3),
  a_f(c(2, 3)) == c(1, 3))

## [1] TRUE

Since a and p generate S3 we deduce that the action is faithfull.

Orbits

S3	Orbits
e	{1_2} {1_3} {2_3}
a	{1_2} {1_3, 2_3}
b	{1_2, 2_3} {1_3}
c	{1_2, 1_3} {2, 3}
p	{1_2, 2_3, 1_3}
p2	{1_2, 2_3, 1_3}

Stabilizers

Action on the elements of order 2

The elements of order 2 are: a, b, c

c(a^p == c, a^(p^2) == b)

## [1] TRUE TRUE

So the action is transitive. And |O(a)|.|Stab(a)| = 6 => 3.|Stab| = 6 So the stabilizer size is 2.

And:

S3	Stab
a	e, a
b	e, b
c	e, c

Action on the elements of order 3

The elements of order 3 are: p, p^2

c(p^a == p^2)

## [1] TRUE

So the action is transitive and: |O(p)|.|Stab(p)| = 6 so |Stab(p)| = 2

S3	Stab
p	e, p, p2
p2	e, p, p2

Other actions

• The action of S3 on S3/<p> is not faithfull. S3/<p> = {{e, p, p2}, {a, b, c}} Any element of <p> is sent to the identity. Any other element is sent to ({e, p, p2} {a, b, c})

• There is also an action of S3 on {a, b, c}.

• And an action of S3 on <p>

All the transitive, faithfull actions of S3

Let \(\phi\) be a transitive, faithfull set action \(S3 \rightarrow S(\mathcal{A})\). Where \(\mathcal{A}\) is any set. We note: \(g \bullet x = \phi(g)(x)\) for \(g \in S3, a \in \mathcal{A}\)

First we would like to show that \(\mathcal{A}\) has 6 elements.

Let \(a_0 \in \mathcal{A}\), we let \(f_{a_0}: S3 \rightarrow \mathcal{A}\) defined by \(f_{a_0}(g) = g \bullet a_0\)

Since the action is transitive \(f_{a_0}\) is surjective. Let’s \(g_1, g_2 \in \mathrm{S3}\). \[f_{a_0}(g_1) = f_{a_0}(g_2) \iff g_1 \bullet a_0 = g_2 \bullet a_0 \iff g_1 g_2^{-1} \bullet a_0 = a_0\]

How to see if 2 actions are isomorphic?

In all this file G is a finite group and S is a set.

An action is a group morphism \(\phi_1: G \rightarrow Aut(S)\) where S is some set. Let \(\phi_2: G \rightarrow Aut(S2)\) be another group action. 2 action are isomorphic iff there exist a set isomorphism \(f: S \rightarrow S2\) such that \[\forall g \in G, f(\phi_1(g)(x)) = \phi_2(g)(f(x))\]

The next property shows that if we have verified the property on f for \(g_1, g_2 \in G\), then the property holds for \(g_1g_2\). \[f(\phi_1(g_1g_2)(x)) = f(\phi_1(g_1)(\phi_1(g_2)(x))) = \phi_2(g_1)(f(\phi_1(g_2)x)) = \phi_2(g_1)(\phi_2(g_2)(fx)) = \phi_2(g_1g_2)(fx)\]

That means that we only need to verify the property on a set of generators for G.

Example

Let’s show that the action on {1, 2, 3} is isomorphic to the action on {{1, 2}, {1, 3}, {2, 3}} by taking f: 1 -> {2, 3} f: 2 -> {1, 3} f: 3 -> {1, 2}

f(p.1) = f(2) = {1, 3} p.f(1) = p.{2, 3} = {1, 3}

f(p.2) = f(3) = {1, 2} p.f(2) = p.{1, 3} = {1, 2}

f(a.1) = f(2) = {1, 3} a.f(1) = a.{2, 3} = {1, 3}

f(a.3) = f(3) = a.f(3) => f(3) = {1, 2}

f(b.2) = f(2) = b.f(2) => f(2) = {1, 3}

f(c.1) = f(1) = c.f(1) => f(1) = {2, 3}

So f is an isomorphism of actions.

Automorphism group of S3

Inner automorphisms

Conjugation by an element creates an automorphism. The conjugation by p has the following effect:

## a^p: (2,3)
## b^p: (1,2)
## c^p: (1,3)
## p^p: (1,2,3)
## p2^p: (1,3,2)

So conj(p) = (a c b)

Here is the conjugation by p2:

## a^p2: (1,3)
## b^p2: (2,3)
## c^p2: (1,2)
## p^p2: (1,2,3)
## p2^p2: (1,3,2)

So conj(p2) = (a b c)

Here is the conjugation by a:

## a^a: (1,2)
## b^a: (2,3)
## c^a: (1,3)
## p^a: (1,3,2)
## p2^a: (1,2,3)

So conj(a) = (b c) (p p2)

Here is the conjugation by b:

## a^b: (2,3)
## b^b: (1,3)
## c^b: (1,2)
## p^b: (1,3,2)
## p2^b: (1,2,3)

So conj(b) = (a c) (p p2)

Here is the conjugation by c:

## a^c: (1,3)
## b^c: (1,2)
## c^c: (2,3)
## p^c: (1,3,2)
## p2^c: (1,2,3)

So conj(c) = (a b)(p p2)

So together with the identity there are 6 inner automorphisms.

Structure of the automorphism group

We will show that Inner(S3) = S3 conj(a)^2 = e conj(p)^3 = e conj(p)^conj(a) = conj a-1 o conj p o conj a = conj a-1pa = conj p2 = conj(p)^2

Since the generators of Inner(S3) have the same relations as those of S3 we know the 2 groups are isomorphic.

Total group or isomorphism

We can show that there are no outer isomorphism. Indeed an automorphism maps elements to elements of the same order. There are thus 2 possibilities for mapping p and 3 possibilities for mapping a. Since a and p generate S3 this 2 images will define uniquely a mapping. There are at most 3*2 automorphisms and according to the previous paragraphs we already have 6 inner automorphisms. So all the automorphism are inner.

Matrix representation of an automorphism

Since an automorphism is uniquely defined by its image of the generators a and p we might be able to write them as a matrix. Indeed the reason we are able to write linear operators as matrix is because they are defined by their action on a basis which is a generator set of the whole vector space.

First lets try to write elements of S3 as vectors: we note a = (1, 0) and p = (0, 1) (0, x) * (1, m) = p^x * a * p^m = a * a p^x * a p^m = (1, 2*x+m)

(g1, n1) * (g2, n2) = g1n1g2n2 = g1g2inv(g2)n1g2n2 = g1g2n1^g2n2 = (g1g2, n1^g2*n2)

inv(g1) * (g2, n2) * g1 = inv(g1) * (g2*g1, n2^g1) = (g2^g1, n2^g1) conj(g1) = g1 e e g1